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The dual of a polyhedron is a polyhedron where the vertices of one correspond to the faces of the other, and vice versa. Is there always a similar correspondence between a pair of polyhedra where the edges of one correspond to the faces of the other, and if so does that relationship have a name? (I'm calling it the edge-face dual of a polyhedron for now.)

Additionally, does this relationship exist between the cube and the rhombic dodecahedron? It's my best guess: cubes have 12 edges and all dodecahedra have 12 faces, and the rhombic dodecahedron is face-transitive and tessellates $\mathbb{R}^3$. If this is the case, then the edge-face dual relationship that I'm looking for is obviously not symmetric, since dodecahedra have 24 edges while cubes only have 6 faces.

Here's some background for what I'm looking for:

One way to extend a hexagonal tiling of $\mathbb{R}^2$ into $\mathbb{R}^3$ is to extrude the hexagon along the new dimension, producing a tessellation of $\mathbb{R}^3$ into hexagonal prisms. I'm looking for a different tiling—one that is related to the isometric projection of a cube, using the observation that a cube is projected into a hexagon. The six adjacent hexagons are connected by six of the cube's projected edges. If you consider the cubes as also tessellating $\mathbb{R}^3$, this leaves six more edges; three "above" and three "below." There are six more cells connected symmetrically to this one. It's not terribly difficult to picture this as being an 3D analogue to a hexagonal grid.

The trouble is, I find it awkward that every connection is between edges, not between faces, and also that each edge touches four other cells, but I'm only considering one to be "adjacent." I find the tiling I get from the above scheme to be desirable; if you're looking at a plane of cells such that you see a hexagonal grid, each cell has six neighbors in that plane, as well as three above (centered on the cell's center) and three below (also centered). I'm trying to find a polyhedron that tessellates $\mathbb{R}^3$ that "looks like" a hexagon in a 2D plane, but where I can consider the adjacent cells to be polyhedra that share faces, and where each plane "stacked" on top of each other is regularly translated (in the same manner as the above scheme).

I can't quite tell if a rhombic dodecahedron is what I want, and if the above description is at all rigorous. This is the only visual aid I've found (of a rhombic dodecahedron), but I'm having a hard time as seeing it as anything other than tiled cubes:

Rhombic dodecahedral honeycomb

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To the question in your title: yes, if you define "edge-face dual" to be the dual of the rectified form of the polyhedron; this seems to have all the properties you might want. (I should expand this into an answer when I have more time, but if anyone else wants to do so, go right ahead...) –  Rahul Jun 30 '11 at 3:25

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Here is an expanded version of my comment.

The rectified form of a polyhedron is a new polyhedron whose vertices lie at the midpoints of the edges of the original one. If you take the dual of this, you obtain a polyhedron whose faces correspond to the edges of the original polyhedron. For example, rectification of a cube yields a cuboctahedron, whose dual is indeed the rhombic dodecahedron. You can see the cube in the rhombic dodecahedron by drawing the short diagonals on all its faces. It is face-transitive because its faces correspond to the edges of the cube, but it is not vertex-transitive because it has two kinds of vertices: those corresponding to the vertices of the cube, and those corresponding to its faces.

I'm pretty sure this is the concept you are looking for, though I wouldn't call it an "edge-face dual" because duality is usually expected to be (close to) an involution. Note that it is the same for a polyhedron and its dual, which we expect because their edges correspond to each other, and which is true because the rectified form of both is the same.

For the rest of the Platonic solids:

  • From a tetrahedron, you get a cube, via an octahedron.
  • From an octahedron, you again get a rhombic dodecahedron, via the cuboctahedron.
  • From a icosahedron or the dodecahedron, you get a rhombic triacontahedron, via the icosidodecahedron.

I can't comment on the rest of your question, as I couldn't follow you to the "3D analogue to a hexagonal grid." However, I can point out that the rhombic dodecahedron is the shape of the Voronoi cell of a face-centered cubic lattice, which does have hexagonal symmetry in an isometric projection, so maybe you're on to something there.

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Thanks. I think the rhombic dodecahedron is what I asked for, but not actually what I want. I'm probably gonna have to give up face-transitivity in the end. –  John Calsbeek Jun 30 '11 at 21:39
    
@John Calsbeek I think the face-centred cubic lattice really is what you are looking for. The rhombic dodecahedron is what you get when you project a tesseract (4D cube) orthogonally along a diagonal into $\mathbb{R}^3$. FCC is a close packing lattice in $\mathbb{R}^3$, as the hex lattice is in the plane. –  yasmar Jul 13 '11 at 11:16

Here are some thoughts formulated in terms of graph theory, and restricted to convex 3-dimensional polyhedra so that Steinitz's Theorem (A graph is 3-polytopal if and only if it is planar and 3-connected) can be used.

There are many ways to take a plane embedding (plane graph) and "transform" it to another plane graph where the new plane graph "inherits" nice properties from the original. The best known of these is the geometric dual where the faces of the original graph become the vertices of the new graph. Another very nice such graph is the medial graph. There is a vertex for each edge of the original and two of these vertices are joined by an edge in the new graph if the edges they represent meet at a vertex and share the same face (for the line graph this last condition is not taken into account). The medial graph of a 3-polytopal graph is also 3-polytopal but it is also 4-valent. Also, dual graphs have the same medial graph.

There are many other nice ways to transform plane 3-connected graphs into new related plane graphs. One interesting one that perhaps is of interest for the original question is to "replace" each edge of the original graph with a hexagon. For the fixed plane embedding outline each face (including the infinite face) inside the face (and close to the original face) with a polygon with the same number of sides. Now join each vertex of these outlines to the nearest vertex of the original graph, and erase the edges of the original graph. Some additional remarks about this are here: http://www.york.cuny.edu/~malk/geometricstructures/2011-session7-TC.html

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